At a certain temperature, 0.885 mol of N2, and 2.654 mol of H2 are placed in a container.
N2+3H2=2NH3
At equilibrium, there is 0.823 mol of NH3 present. Determine the number of mol of N2 and H2 that are present when the reaction is at equilibrium.
Answer in moled of N2 at equilibrium and moles of H2 at equilibrium.
To determine the number of moles of N2 and H2 at equilibrium, we can use the stoichiometry of the balanced chemical equation. The balanced equation is N2 + 3H2 = 2NH3.
First, let's calculate the number of moles of NH3 produced at equilibrium. We are given that at equilibrium, there is 0.823 mol of NH3 present.
Since the stoichiometry of the balanced equation is 1:3:2 for N2:H2:NH3, this means that for every 2 moles of NH3 produced, we need 1 mole of N2 and 3 moles of H2.
Therefore, using a ratio, we can calculate the moles of N2 and H2 at equilibrium:
Moles of N2 at equilibrium = (0.823 mol NH3) x (1 mol N2 / 2 mol NH3)
Moles of H2 at equilibrium = (0.823 mol NH3) x (3 mol H2 / 2 mol NH3)
Performing the calculations:
Moles of N2 at equilibrium = 0.823 mol NH3 / 2 = 0.4115 mol N2
Moles of H2 at equilibrium = 0.823 mol NH3 x 3 = 2.469 mol H2 / 2 = 1.2345 mol H2
Therefore, at equilibrium, there are approximately 0.4115 moles of N2 and 1.2345 moles of H2 present.
well, you have 3 mol of H2 for every mol of N2 so your reaction is not limited by either
so
You have the same number of atoms of N2 on left as on right so
.823/.885 = .93 of each reacted
so .07 of each remains
.07 * .885 = .062 Mol of N2 still there
and
.07 * 2.654 = .186 mol of H2 remains